# Proof for an easy empirical result on fractions and ceilings

Claim: The fractional part of $$\frac{2z}{i}$$, where $$z$$ and $$i$$ are finite is no more than $$\frac{i-2}{i}$$ when $$i$$ is even and no more than $$\frac{i-1}{i}$$ when $$i$$ is odd. Here $$z,i$$ are non-negative integers. To make the problem non-trivial, assume $$i \ge 3$$.

Example: If $$i=4$$, then $$2z$$ can be $$0,2,4,6,...12...$$. The fractional parts for these numbers then are $$0,2/4$$. The largest of this is 2/4 which is $$\frac{i-2}{i}$$.

Second example: If $$i=7$$, then $$2z$$ can be $$0,2,4,6,...42...$$. The fractional parts for these numbers then are $$0,2/7, 4/6,6/7,3/7,5/7$$. The largest of this is 6/7 which is $$\frac{i-1}{i}$$.

I cannot write this simple result in the form of a rigorous proof. How should I?

UPDATE: I have provided an answer myself. Corrections are welcome.

• Why has this been downvoted without any explanation? Jan 24, 2020 at 14:33

If $$b$$ is even, $$\{\frac{2a}{b} \}$$ is one of $$\{ 0, \frac{2}{b}, \frac{4}{b}, \frac{6}{b}, \ldots, ... \frac{b-2}{b} \}$$. The maximum of these is $$\frac{b-2}{b}$$. If $$b$$ is odd, $$\{\frac{2a}{b} \}$$ is one of $$\{ 0, \frac{2}{b}, \frac{4}{b}, \frac{6}{b}, \ldots, ... \frac{b-1}{b} \}$$. The maximum of these is $$\frac{b-1}{b}$$.